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andre [41]
3 years ago
14

The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quan

tity to fall to half its original amount. Carbon 14 has a half-life of 5,730 years. Suppose given samples of carbon 14 weigh (fraction 5/8) of a pound and (fraction 7/8) of a pound. What was the total weight of the samples 11460 years ago
(show work please)
Mathematics
2 answers:
Ilya [14]3 years ago
6 0
That's two half lives. 5/8 * 2 * 2 = 20/8 = 5/2 pounds7/8 * 2 * 2 = 28/8 = 7/2 pounds
Drupady [299]3 years ago
3 0

Answer:

Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.

Step-by-step explanation:

Radioactive decay is an exponential process represented by

A_{t}=A_{0}e^{-kt}

where A_{t} = Amount of the radioactive element after t years

A_{0} = Initial amount

k = Decay constant

t = time in years

Half life period of Carbon-14 is 5730 years.

\frac{A_{0} }{2}=A_{0}e^{-5730k}

\frac{1}{2}=e^{-5730k}

Now we take ln (Natural log) on both the sides

ln(\frac{1}{2})=ln[e^{-5730k}]

-ln(2) = -5730kln(e)

0.69315 = 5730k

k=\frac{0.69315}{5730}

k=1.21\times 10^{-4}

Now we have to calculate the weight of samples of C-14 taken for the remaining quantities \frac{5}{8} and \frac{7}{8} of a pound.

\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}

\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}

\frac{5}{8}=A_{0}e^{(-1.3863)}

A_{0}=\frac{5}{8}\times e^{1.3863}

A_{0}=\frac{5}{8}\times 4

A_{0}=\frac{5}{2}

A_{0}=2.5 pounds

Similarly for \frac{7}{8} pounds

\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}

\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}

\frac{7}{8}=A_{0}e^{(-1.3863)}

A_{0}=\frac{7}{8}\times e^{(1.3863)}

A_{0}=\frac{7}{8}\times 4

A_{0}=\frac{7}{2}

A_{0}=3.5 pounds

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